Count number of pairs (A <= N, B <= N) such that gcd (A , B) is B

Given a number n. We need to find the number of ordered pairs of a and b such gcd(a, b) is b itself
Examples :

Input : n = 2
Output : 3
(1, 1) (2, 2) and (2, 1) 

Input : n = 3
Output : 5
(1, 1) (2, 2) (3, 3) (2, 1) and (3, 1)

Naive approach: gcd(a, b) = b means b is a factor of a. So the total number of pairs will be equal to sum of divisors for each a = 1 to n. Please refer find all divisors of a natural number for implementation.
Efficient approach: gcd(a, b) = b means that a is a multiple of b. So the total number of pairs will be sum of number of multiples of each b (where b varies from 1 to n) which are less than or equal to n. 
For a number i, a number of multiples of i is less than or equal to floor(n/i). So what we need to do is just sum the floor(n/i) for each i = 1 to n and print it. But more optimizations can be done. floor(n/i) can have atmost 2*sqrt(n) values for i >= sqrt(n). floor(n/i) can vary from 1 to sqrt(n) and similarly for i = 1 to sqrt(n) floor(n/i) can have values from 1 to sqrt(n). So total of 2*sqrt(n) distinct values 

let floor(n/i) = k
k <= n/i < k + 1
n/k+1 < i <= n/k
floor(n/k+1) < i <= floor(n/k)
Thus for given k the largest value of i for 
which the floor(n/i) = k is floor(n/k)
and all the set of i for which the 
floor(n/i) = k are consecutive

CPP

// C++ implementation of counting pairs
// such that gcd (a, b) = b
#include <bits/stdc++.h>
using namespace std;

// returns number of valid pairs
int CountPairs(int n)
{
    // initialize k
    int k = n;

    // loop till imin <= n
    int imin = 1;

    // Initialize result
    int ans = 0;

    while (imin <= n) {

        // max i with given k floor(n/k)
        int imax = n / k;

        // adding k*(number of i with
        // floor(n/i) = k to ans
        ans += k * (imax - imin + 1);

        // set imin = imax + 1 and k = n/imin
        imin = imax + 1;
        k = n / imin;
    }

    return ans;
}

// Driver function
int main()
{
    cout << CountPairs(1) << endl;
    cout << CountPairs(2) << endl;
    cout << CountPairs(3) << endl;
    return 0;
}

Java

// Java implementation of counting pairs
// such that gcd (a, b) = b
import java.io.*;
public class GFG {

    // returns number of valid pairs
    static int CountPairs(int n) {

        // initialize k
        int k = n;

        // loop till imin <= n
        int imin = 1;

        // Initialize result
        int ans = 0;

        while (imin <= n) {

            // max i with given k floor(n/k)
            int imax = n / k;

            // adding k*(number of i with
            // floor(n/i) = k to ans
            ans += k * (imax - imin + 1);

            // set imin = imax + 1
            // and k = n/imin
            imin = imax + 1;
            k = n / imin;
        }

        return ans;
    }

    // Driver code
    public static void main(String[] args) {
        System.out.println(CountPairs(1));
        System.out.println(CountPairs(2));
        System.out.println(CountPairs(3));
    }
}

// This code is contributed by Anant Agarwal.

Python3

# Python implementation of counting
# pairs such that gcd (a, b) = b

# returns number of valid pairs
def CountPairs(n):

    # initialize k
    k = n

    # loop till imin <= n
    imin = 1

    # Initialize result
    ans = 0

    while(imin <= n):

        # max i with given k floor(n / k)
        imax = n / k

        # adding k*(number of i with
        # floor(n / i) = k to ans
        ans += k * (imax - imin + 1)

        # set imin = imax + 1 and
        # k = n / imin
        imin = imax + 1
        k = n / imin

    return ans

# Driver code
print(CountPairs(1))
print(CountPairs(2))
print(CountPairs(3))

# This code is contributed by Anant Agarwal.

C#

// C# implementation of counting
// pairs such that gcd (a, b) = b
using System;

class GFG {

    // returns number of valid pairs
    static int CountPairs(int n)
    {

        // initialize k
        int k = n;

        // loop till imin <= n
        int imin = 1;

        // Initialize result
        int ans = 0;

        while (imin <= n) {

            // max i with given
            // k floor(n / k)
            int imax = n / k;

            // adding k * (number of i
            // with floor(n / i) = k
            // to ans
            ans += k * (imax - imin + 1);

            // set imin = imax + 1
            // and k = n / imin
            imin = imax + 1;
            k = n / imin;
        }

        return ans;
    }

    // Driver code
    public static void Main(String []args)
    {
        Console.WriteLine(CountPairs(1));
        Console.WriteLine(CountPairs(2));
        Console.WriteLine(CountPairs(3));
    }
}

// This code is contributed by vt_m.

PHP

<?php
// PHP implementation of counting
// pairs such that gcd (a, b) = b

// returns number of valid pairs
function CountPairs($n)
{

    // initialize k
    $k = $n;

    // loop till imin <= n
    $imin = 1;

    // Initialize result
    $ans = 0;

    while ($imin <= $n)
    {

        // max i with given k floor(n/k)
        $imax = $n / $k;

        // adding k*(number of i with
        // floor(n/i) = k to ans
        $ans += $k * ($imax - $imin + 1);

        // set imin = imax + 1
        // and k = n/imin
        $imin = $imax + 1;
        $k = (int)($n / $imin);
    }

    return $ans;
}

// Driver Code
echo(CountPairs(1) . "\n");
echo(CountPairs(2) . "\n");
echo(CountPairs(3) . "\n");

// This code is contributed by Ajit.
?>

Javascript

<script>

// Javascript implementation of counting pairs
// such that gcd (a, b) = b

// returns number of valid pairs
function CountPairs(n)
{
    // initialize k
    let k = n;

    // loop till imin <= n
    let imin = 1;

    // Initialize result
    let ans = 0;

    while (imin <= n) {

        // max i with given k floor(n/k)
        let imax = Math.floor(n / k);

        // adding k*(number of i with
        // floor(n/i) = k to ans
        ans += k * (imax - imin + 1);

        // set imin = imax + 1 and k = n/imin
        imin = imax + 1;
        k = Math.floor(n / imin);
    }

    return ans;
}

// Driver function

    document.write(CountPairs(1) + "<br>");
    document.write(CountPairs(2) + "<br>");
    document.write(CountPairs(3) + "<br>");

// This is code is contributed by Mayank Tyagi

</script>

Output

1
3
5

Time complexity: O(n). This is because the while loop takes O(n) time to complete since it is looping over all elements of the array. 
Auxiliary space: O(1), as no extra space is used.

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